Question

Can someone help me with this

Answer 1

19 m

Explanation:

This is Geometric progression with

`\left \{ {{u_1=1,\: u_2 =2} \atop {u_n=2^(n-1)u_1} \right.`

Then the fomular of sum is

`S=u_1(2^n-1)/(2-1) =1*(2^n-1)=2^n-1\\To\: save\: 1 \:million\: then \; our \:sum \: will \: be\: 1000000\\Hence,\\2^n-1=1000000\\2^n=1000001\\n=log_(2){1000001}\approx19,9\\Thus, it\:would \;be \:19 \:months \:before\: we\: saved \:1000000 \:pounds`